Respuesta :
[tex]\bf (\stackrel{x_1}{4}~,~\stackrel{y_1}{-6})\qquad (\stackrel{x_2}{8}~,~\stackrel{y_2}{-4}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-4}-\stackrel{y1}{(-6)}}}{\underset{run} {\underset{x_2}{8}-\underset{x_1}{4}}}\implies \cfrac{-4+6}{4}\implies \cfrac{2}{4}\implies \cfrac{1}{2}[/tex]
[tex]\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-6)}=\stackrel{m}{\cfrac{1}{2}}(x-\stackrel{x_1}{4}) \\\\\\ y+6=\cfrac{1}{2}x-2\implies y=\cfrac{1}{2}x-8[/tex]
Answer:
y = 1/2x - 8.
Step-by-step explanation:
The slope of the line = rise / run
= (-4 - (-6)) / (8 - 4)
= 2/4
= 1/2.
Using the point/slope form of the equation of a line:
y - y1 = m(x - x1)
Here we have m = 1/2 and (x1, y1) = (4, -6):
y - (-6) = 1/2 (x - 4)
y + 6 = 1/2x - 2
y = 1/2x - 8.
